Differentiation (Part-38) | S N De

 

$\,3.\,$ If $\,\sin\alpha+\sin\beta=a,\,\,\cos\alpha+\cos\beta=b,\,$ find the value of $\,\,\cos(\alpha+\beta).$

Sol.  $\,\,\,\,\,\frac ab= \frac{\sin\alpha+\sin\beta}{\cos\alpha+\cos\beta} \\ \Rightarrow \frac ab=\frac{2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}}{2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}}\\~~~~~~~~=\tan\frac{\alpha+\beta}{2} \\ \therefore\cos(\alpha+\beta)=\frac{1-\tan^2\frac{\alpha+\beta}{2}}{1+\tan^2\frac{\alpha+\beta}{2}}\\~~~~~~~~~~~~~~~~~=\frac{1-\frac{a^2}{b^2}}{1+\frac{a^2}{b^2}} \\~~~~~~~~~~~~~~~~~=\frac{b^2-a^2}{b^2+a^2}$

$\,4.\,$ If $\,\,\sec(\phi+\alpha)+\sec(\phi-\alpha)=2\sec\phi,\,\,$ prove that, $\,\,\cos\phi=\pm\sqrt2\cos\frac{\alpha}{2}$

Sol. $\,\,\sec(\phi+\alpha)+\sec(\phi-\alpha)=2\sec\phi \\ \Rightarrow \frac{1}{\cos(\phi+\alpha)}+\frac{1}{\cos(\phi-\alpha)}=\frac{2}{\cos\phi} \\ \Rightarrow \frac{\cos(\phi+\alpha)+\cos(\phi-\alpha)}{\cos(\phi+\alpha)\cos(\phi-\alpha)}=\frac{2}{\cos\phi} \\ \Rightarrow \frac{2\cos\phi\cos\alpha}{\frac 12(\cos2\phi+\cos2\alpha)}=\frac{2}{\cos\phi} \\ \Rightarrow \frac{\cos\phi\cos\alpha}{\frac 12(\cos2\phi+\cos2\alpha)}=\frac{1}{\cos\phi} \\ \Rightarrow 2\cos^2\phi \cos\alpha=\cos2\phi+\cos2\alpha \\ \Rightarrow 2\cos^2\phi \cos\alpha=2\cos^2\phi-1+1-2\sin^2\alpha \\ \Rightarrow 2\cos^2\phi(\cos\alpha-1)=-1+1-2\sin^2\alpha \\ \Rightarrow \cos^2\phi=\frac{-\sin^2\alpha}{\cos\alpha-1}\\ ~~~~~~~~~~~~~~~=\frac{\sin^2\alpha}{1-\cos\alpha}\\~~~~~~~~~~~~~~~=\frac{1-\cos^2\alpha}{1-\cos\alpha}\\~~~~~~~~~~~~~~~=\frac{(1+\cos\alpha)(1-\cos\alpha)}{1-\cos\alpha}\\~~~~~~~~~~~~~~~=1+\cos\alpha\\~~~~~~~~~~~~~~~=2\cos^2\frac{\alpha}{2}\\ \Rightarrow \cos\phi=\pm \sqrt2 \cos\alpha$

$\,5.\,$ If $\,\,\cos\theta=\frac{\cos\alpha-\cos\beta}{1-\cos\alpha\cos\beta},\,\,$ show that, one value of $\,\,\tan\frac{\theta}{2}\,\,$ is $\,\,\tan\frac{\alpha}{2}\cot\frac{\beta}{2}.$

Sol. $\,\,\cos\theta=\frac{\cos\alpha-\cos\beta}{1-\cos\alpha\cos\beta}\\ \Rightarrow \frac{1-\cos\theta}{1+\cos\theta}=\frac{1-\cos\alpha\cos\beta-\cos\alpha+\cos\beta}{1-\cos\alpha\cos\beta+\cos\alpha-\cos\beta}\\ \Rightarrow \tan^2\frac{\theta}{2}=\frac{(1-\cos\alpha)(1+\cos\beta)}{(1-\cos\beta)(1+\cos\alpha)}\\~~~~~~~~~~~~~~~~=\frac{2\sin^2\frac{\alpha}{2}2\cos^2\frac{\beta}{2}}{2\sin^2\frac{\beta}{2}2\cos^2\frac{\alpha}{2}} \\ \Rightarrow \tan^2\frac{\theta}{2}=\tan^2\frac{\alpha}{2}\cot^2\frac{\beta}{2} \\ \Rightarrow \tan\frac{\theta}{2}=\pm \tan\frac{\alpha}{2}\cot\frac{\beta}{2}$

Hence,  one value of $\,\,\tan\frac{\theta}{2}\,\,$ is $\,\,\tan\frac{\alpha}{2}\cot\frac{\beta}{2}.$

$\,6.\,$ If $\,\,A=170^{\circ},\,\,$ prove that , $\,\tan\frac{A}{2}=\frac{-1-\sqrt{1+\tan^2A}}{\tan A}$

Sol.$\,\,\,\frac{-1-\sqrt{1+\tan^2A}}{\tan A}\\=\frac{-1-\sqrt{\sec^2A}}{\tan A}\\=\frac{-1-(-\sec A)}{\tan A}\,\,\,\,[\text{since,}\,\,\sec 170^{\circ}<0]\\=\frac{-1+\sec A}{\tan A}\\=\frac{1-\cos A}{\sin A}\\=\frac{2\sin^2\frac A2}{2\sin\frac A2\cos \frac A2}\\=\frac{\sin \frac A2}{\cos\frac A2}\\=\tan \frac A2$

$\,7.\,$ Prove that , $\,\,2\sin \frac A2=\pm \sqrt{1+\sin A}\pm \sqrt{1-\sin A}.$ Hence, determine the correct signs of the R.H.S  when $\,180^{\circ} <A<270^{\circ}.$

Sol.  $\,\,1+\sin A= \sin^2\frac A2+\cos^2\frac A2+2\sin\frac A2\cos \frac A2 \\~~~~~~~~~~~~~~~~~=(\sin \frac A2+\cos\frac A2)^2 \\ \Rightarrow \pm \sqrt{1+\sin A}=\sin\frac A2+\cos\frac A2 \cdots(1)$

Similarly, $\,\,11\sin A= \sin^2\frac A2+\cos^2\frac A2-2\sin\frac A2\cos \frac A2 \\~~~~~~~~~~~~~~~~~=(\sin \frac A2-\cos\frac A2)^2 \\ \Rightarrow \pm \sqrt{1-\sin A}=\sin\frac A2-\cos\frac A2 \cdots(2)$

Adding (1) and (2), we get $\,\,2\sin \frac A2=\pm \sqrt{1+\sin A}\pm \sqrt{1-\sin A}.$

2nd part : Since $\,\,180^{\circ}< A< 270^{\circ} \\ \Rightarrow \frac{180^{\circ}}{2}<\frac A2<\frac{270^{\circ}}{2} \\ \Rightarrow 90^{\circ}<\frac A2<135^{\circ} \\ \Rightarrow 90^{\circ}+45^{\circ}<\frac A2+45^{\circ}<135^{\circ}+45^{\circ} \\ \Rightarrow 135^{\circ}<\frac A2+45^{\circ}<180^{\circ} \cdots(1) \\ \therefore \,\, \sin\frac A2+\cos\frac A2 \\=\sqrt2(\frac{1}{\sqrt2}\sin\frac A2+\frac{1}{\sqrt2}\cos \frac A2 )\\=\sqrt2\sin(\frac A2+45^{\circ})>0\,\,\,\,[\text{By (1)}]$

Again, since $\,\,90^{\circ}<\frac A2<135^{\circ},\,\,\sin\frac A2>0,\,\cos \frac A2<0 \\ \text{so,}\,\,\sin \frac A2-\cos\frac A2>0.$

Hence, $\,\,\sin \frac A2+\cos\frac A2=\sqrt{1+\sin A},\\ \sin\frac A2-\cos\frac A2=\sqrt{1-\sin A}. \\ \therefore 2\sin\frac A2=\sqrt{1+\sin A}+\sqrt{1-\sin A}$

$\,8.\,$ Given $\,\theta=-320^{\circ},\,\,$ find the correct sign of the following identity : $\,\,\tan \frac{\theta}{2}=\frac{-1\pm \sqrt{1+\tan^2\theta}}{\tan \theta}$

Sol. We compute,$\,\,\frac{-1+\sqrt{1+\tan^2\theta}}{\tan \theta}\\=\frac{-1+\sqrt{\sec^2\theta}}{\tan\theta} \cdots(1)\\=\frac{-1+\sec\theta}{\tan\theta}\\=\frac{1-\cos\theta}{\sin\theta}\\=\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\\=\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}\\=\tan\frac{\theta}{2}$

 Similarly, $\,\,\frac{-1-\sqrt{1+\tan^2\theta}}{\tan \theta}\\=\frac{-1-\sqrt{sec^2\theta}}{\tan\theta}\\=\frac{-1-\sec\theta}{\tan\theta}\\=\frac{-(1+\sec\theta)}{\tan\theta}\\=\frac{-(1+\cos\theta)}{\sin\theta}\\=\frac{-2\cos^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)}\\=-\cot(\theta/2)$

Again, alternatively, for $\,\,\theta=-320^{\circ},\\\pm\sqrt{1+\tan^2\theta}=+\sqrt{1+\tan^2\theta}=\sec\theta$ 

and we can directly use (1).

Hence, the correct sign of the following identity is : $\,\,\tan \frac{\theta}{2}=\frac{-1+\sqrt{1+\tan^2\theta}}{\tan \theta}$

$\,9.\,$  Prove that, $\,\,\sin 5^{\circ}-\sin 67^{\circ}+\sin 77^{\circ}-\sin 139^{\circ}+\sin 149^{\circ}=0$

Sol. $\,\,\sin 5^{\circ}-\sin 67^{\circ}+\sin 77^{\circ}-\sin 139^{\circ}+\sin 149^{\circ}\\=\sin 5^{\circ}+(\sin 77^{\circ}-\sin 67^{\circ})\\+(\sin 149^{\circ}-\sin 139^{\circ})\\=\sin 5^{\circ}+2\cos\frac{77^{\circ}+67^{\circ}}{2}\sin\frac{77^{\circ}-67^{\circ}}{2}\\+2\cos\frac{149^{\circ}+139^{\circ}}{2}\sin\frac{149^{\circ}-139^{\circ}}{2}\\=\sin 5^{\circ}+2\sin 5^{\circ}\cos 72^{\circ}\\+2\sin 5^{\circ}\cos 144^{\circ}\\=\sin 5^{\circ}+2\sin 5^{\circ}\cos72^{\circ}\\+2\sin 5^{\circ}\cos(90^{\circ}+54^{\circ})\\=\sin 5^{\circ}+2\sin 5^{\circ}\sin(90^{\circ}-72^{\circ})\\+2\sin 5^{\circ}(-\sin 54^{\circ})\\=\sin 5^{\circ}+2\sin 5^{\circ}\sin 18^{\circ}-2\sin 5^{\circ}\sin 54^{\circ}\\=\sin 5^{\circ}-2\sin5^{\circ}(\sin 54^{\circ}-\sin 18^{\circ})\\=\sin 5^{\circ}-2\sin 5^{\circ}. 2\cos\frac{54^{\circ}+18^{\circ}}{2}\sin\frac{54^{\circ}-18^{\circ}}{2}\\=\sin 5^{\circ}-2\sin 5^{\circ}. 2\cos 36^{\circ}\sin 18^{\circ}\\=\sin 5^{\circ}-2\sin 5^{\circ}.2\times \frac{\sqrt5+1}{4}\times \frac{\sqrt5-1}{4}\\=\sin 5^{\circ}-\sin 5^{\circ}. \left[2 \times 2\times \frac{1}{16}\times  (5-1)\right]\\=\sin 5^{\circ}-\sin 5^{\circ}\\=0$